 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/5/solution

F = (cos (y^{5}) sinh z, - 5 x y^{4} sin (y^{5}) sinh z, x cos (y^{5}) cosh z) .

The factor

5 y^{4}

in the middle component comes from the chain rule. The potential is smooth, so the curl of a gradient identity gives

\nabla \times F = 0

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